# Evaluate $\lim_{x \to 0} \dfrac{\sin(2x)}{8x}$

I need to evaluate $$\lim_{x \to 0} \dfrac{\sin(2x)}{8x}$$ I'm doing Calculus One course on coursera.org and here is their explanation that doesn't explain anything for me:

Why $\lim_{x \to 0} \dfrac{\sin(x)}{x}=1$ ?

Why do I need to set $y=2x$? And why $y\to 0 as x\to 0$?

Why $\lim_{y \to 0} \dfrac{\sin(y)}{y} \cdot \dfrac{2}{8}$ suddenly became $\lim_{x \to 0} \dfrac{\sin(2x)}{8x}=\dfrac{1}{4}$?

• if you have you a calculator that can compute $\sin?$ then compute $\frac{\sin x}{x}$ for $x = 0.1, 0.01, 0.001$ and see what happens. look at the graphs of $\sin x$ and $x$ in the window $-1 \le x \le 1, -1 \le y \le 1$
– abel
Feb 24, 2015 at 12:50

For the first one, why not google it?
Then $$\lim_{x\to 0}\frac{\sin 2x}{8x}=\lim_{2x\to 0}\frac{\sin 2x}{4\cdot 2x}=\frac14\lim_{2x\to 0}\frac{\sin 2x}{2x}=\frac14\cdot 1=\frac14$$

• Sorry, I don't understand anything. Why 8x became 2x? WHY is this whole equation equal to $1/4$? Feb 24, 2015 at 12:06
• @mega_charizard I've updated my answer, you can see it.
– Vim
Feb 24, 2015 at 12:08
• Now I don't understand even more than before. Why you change x-->0 to 2x-->0? Feb 24, 2015 at 12:16
• @mega_charizard Coz they're equivalent. When $x\to0$ doesn't $2x$ also $\to 0$?
– Vim
Feb 24, 2015 at 12:17
• Ok, I think I get why changing $x\to0$ to $2x\to0$ helps to solve the problem. But I must have missed something in the course, because I don't know why $x\to0$ is the same as $2x\to0$. Feb 24, 2015 at 12:26

$\bf hint:$ if you have you a calculator that can compute $\sin,$ then compute $\frac{\sin x}{x}$ for $x = 0.1, 0.01, 0.001$ and see what happens. look at the graphs of $\sin x$ and $x$ in the window $-1 \le x \le 1, -1 \le y \le 1$ make sure your calculator is in radian mode. happy experimenting.